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a sample containing only carbon, hydrogen, phosphorus, and oxygen is subjected to elemental analysis. after complete combustion, a 0.3408-g sample of the compound yields 0.4887 g of co2, 0.3001 g of h2o, and 0.2627 g of p4o10. what is the empirical formula of the compound?

Respuesta :

The empirical formula of compound is C1H3P

Mass of Sample = 0.3408-g

Mass of CO2 = 0.4887g

Mass of H2O = 0.3001 g

Mass of P4O10 = 0.2627 g

First we have to calculate moles of CO₂, H2O and P4O10 formed.

Moles of CO2 = 0.4887g/44 g/mol = 0.011 mol

Now, Moles of carbon == Moles of CO₂ = 0.011 mol

Moles of H2O = 0.3001 g/18 g/mol = 0.016 mol

Now, Moles of hydrogen = 2 x Moles of H2O = 2 × 0.016 = 0.032 mol

Moles of P4O10 = 0.2627 g/283 g/mol = 0.0009 mol

Now, Moles of Phosphorous = Moles of P4O10 = 4 x 0.0009 moles = 0.0036 mol

Therefore, the ratio of number of moles of C: H: P is = 0.011 : 0.032 : 0.0036

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = 0.011/0.011 = 1

For H = 0.032/0.011 = 2.9 = 3

For P = 0.0036/0.011 = 0

Thus, C: H: P = 1 : 3 : 0

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is C1H3P

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